Does a change in direction (turnaround) result in loss of synchronization?

[Tim88]

In the philosophy forum we are discussing interpretations of the car scenario that was earlier discussed in this forum. The car has two synchronized clocks.

 

While rehashing it, I turned it "on the fly" into a twin scenario by letting the car do a turnaround.

To keep it simple, a let the car take a sharp turn without changing speed.

 

Without much reflection I stated that after turnaround the car's "reference system has been messed up: the speed of light doesn't seem the same anymore in both directions" because the clock synchronization was done for a different inertial reference system.

 

Mordred added that "A change in direction results in loss of synchronization (turnaround)"

 

But when looking at it more carefully, I realized that I had based my statement on a different scenario with linear acceleration. And according to my new analysis, our statements were wrong.

 

Here's my analysis, which is very simple, all from the perspective of the ground frame:

 

1. The car's rear clock is advanced on the front clock by a time difference D, such that for the car the speed of light appears isotropic relative to the car.

 

2. In this particular case the clock rates are kept identical and constant, because (neglecting the asymmetry of the car) both clocks keep on moving at the same speed even while taking the turn.

 

3. Accounting for the turnaround, the time difference is still D.

 

4. After taking the turn, the front clock is still in front and the rear clock is still at the rear. It's just the mirror case of the first leg of the trip.

 

In other words, concerning the car we have the same physical situation as before: the car is still moving at the speed v with a clock synchronization that is the standard one for that speed. No resynchronization is required to newly obtain a measured speed of light c in both directions.

 

[Mordred]

Quite frankly the "Under constant velocity term" should be enough to realize the Einstein synchronization procedure is no longer valid under "rapidity".

 

Under rapidity different points on the car itself are undergoing spatial rotation. This changes the time interval due to rotation. Try and distinquish between "time" and "time interval". Minkowskii uses "Intervals" In time interval under Lorentz you have two synchronization equations "outgoing Lorentz" incoming "Lorentz" due to isotropy of time the synchronization transforms are inverse to each other.( Both assume constant velocity). Under rapidity its a 180 degree change that does not detail all the rapidity points between incoming and outgoing. So on spacetime diagrams you have a region not accounted for. Assuming your strictly using the Lorentz transformation base formulas.

The problem with your above is you aren't accounting for "DURING ROTATION" to get from inbound to outbound. In order to maintain synchronization during rotation you must run the transforms with x following a hyperbolic curve. If you were to assign different observer points on the car. A front, B back C centre. The only point that will be valid under rotation to Einstein Synchronization (base) is point C.( the transformations account for x plus and x minus ONLY) *assuming centre of rotation. However both A and B are in different Spatial directions simultaneously.

 

Now even if you just use event C. the above only accounts for vectors at 90 degrees and 180 degrees. Not for all the angles in between.

Remember SR uses VECTORS. Your vectors undergo rotation in which is the spatial component's are affected.

 

As the spatial components are affected your "TIME INTERVAL" is affected.

 

I have one question. The above is detailed in any SR textbook. If your getting answers that says these books are wrong....?

 

Rapidity

 

"Mathematically, rapidity can be defined as the hyperbolic angle that differentiates two frames of reference in relative motion, each frame being associated with distance and time coordinates.

 

Proper acceleration and rapidity.

" (the acceleration 'felt' by the object being accelerated) is the rate of change of rapidity with respect to proper time"

 

under each section look at the transforms on Gamma and Beta components.

 

For example acceleration. [latex]\beta\gamma=sinhW [/latex]

 

https://en.m.wikipedia.org/wiki/Rapidity

Edited October 9, 2016 by Mordred

[Tim88]

Quite frankly the "Under constant velocity term" should be enough to realize the Einstein synchronization procedure is no longer valid under "rapidity".

 

Under rapidity different points on the car itself are undergoing spatial rotation. This changes the time interval due to rotation.

 

Well, that kind of reasoning went in a flash through my head at the moment that I typed my first comment that the car's synchronization will be messed up; it seemed pretty safe to claim that the synchronization would be off. And in most cases that's no doubt correct.

 

 

[..] Assuming your strictly using the Lorentz transformation base formulas. [..]

The problem with your above is you aren't accounting for "DURING ROTATION" to get from inbound to outbound. In order to maintain synchronization during rotation you must run the transforms with x following a hyperbolic curve. [..]

 

No, not at all - I can't understand how you can assume that I made use of complex equations that are unnecessary for the analysis. Instead, as you can read in my first post, I used physical reasoning - just like Einstein did when stating "It is at once apparent" (§4).

 

This is also a bit like Bell's spaceship paradox, as Bell similarly used physical reasoning while his colleagues used complex formula's.

 

So, it will be difficult to find a mistake in your hyperbolic curves or their interpretation if you made one, due to their complexity.

In contrast, if I made a mistake then it should be very easy to spot!

 

If anyone sees a problem with any of the points 1-4, or has questions about them, please tell.

PS I did not see this particular case anywhere else; but of course the literature is huge. Anyway, we won't need it as it's quite simple.

[Mordred]

I already pointed out the errors in 1 to 4. Assumption. You are assuming No rapidity.

Twin Paradox

event one observes event 2, neither A or B can tell which observer is inertial. (Einstein synchronized). So returning twin should be the same age as at home twin. If you assume no Rapidity. Case one Scalar only We can actually solve this without math....

event a watches event b. Yells hey wait a minute. Event B is slowing down/speeding up. ( No longer on the same Worldline).

Second case. Physical direction change=change in rapidity. Observer A yells "Hey wait a minute he is turning".

more complexely shown the worldline ds^2 path also changed.

 

The ds^2 worldline is your synchronized events. When you add rapidity in both cases the Worldline path for synchronized events change. You are no longer on the same spacetime geodesics when you accelerate. Your spacetime diagram assumes constant velocity. Those diagrams set the ds^2 line elements accordingly. Once you have rapidity that same spacetime diagram is no longer valid.

 

Katra. acceleration causes rapidity. Which changes the worldline to a different worldline.

Edited October 9, 2016 by Mordred

[Tim88]

I already pointed out the errors in 1 to 4. Assumption. You are assuming No rapidity.

Twin Paradox

event one observes event 2, neither A or B can tell which observer is inertial. (Einstein synchronized). So returning twin should be the same age as at home twin. If you assume no Rapidity. Case one Scalar only We can actually solve this without math....

event a watches event b. Yells hey wait a minute. Event B is slowing down/speeding up. ( No longer on the same Worldline).

Second case. Physical direction change=change in rapidity. Observer A yells "Hey wait a minute he is turning".

more complexely shown the worldline ds^2 path also changed.

 

The ds^2 worldline is your synchronized events. When you add rapidity in both cases the Worldline path for synchronized events change. You are no longer on the same spacetime geodesics when you accelerate. Your spacetime diagram assumes constant velocity. Those diagrams set the ds^2 line elements accordingly. Once you have rapidity that same spacetime diagram is no longer valid.

 

Katra. acceleration causes rapidity. Which changes the worldline to a different worldline.

 

In the points under discussion:

 

- I made no assumption about "rapidity"; that concept is not used.

- I made no use of wordlines or spacetime diagrams.

 

And neither did Einstein and Bell in the referred places.

Note also that the later following "twin" clock measurements are not affected by eventual 1-way speed measurements of the car driver.

 

What here instantly became clear to me, is effectively a combination of Einstein's and Bell's above-mentioned instant understandings; my reference to those is therefore not only useful as example but was directly applied.

Strangely enough you did not mention any of it in your criticism of my points 1-4...

[Mordred]

Its amazing you constantly ignore what I am describing to you. The last post directly affects your 1 to 4. What is it about the statement. "when you have rapidity, your no longer on the same worldline" didn't you understand?

 

If your on a different worldline you have a different set of simultaneous events.

 

So "Does a change in direction result in a loss of synchronization" ?

 

Absolutely 100% YES. Care to prove me wrong. That is described in any standard textbook.

Edited October 10, 2016 by Mordred

[Tim88]

And where in his analysis of circular motion did Einstein use rapidity? I didn't see you make an analysis that allows an exact prediction of the time difference according to the ground frame, starting from the initial time difference D.

[edit:] Maybe you accept Einstein's result to which I linked in post #3, and which is easy to apply to the two clocks. Starting from that accepted result, calculating backwards (straight line constant speed) you should find the same as me.

 

And as you did not make clear with which step you have problems, let's take it step by step: please clarify the perceived error in my step 1.

Edited October 10, 2016 by Tim88

[swansont]

And where in his analysis of circular motion did Einstein use rapidity?

 

 

Why is this relevant? I see nothing in the OP asking that we limit the discussion to Einstein's analysis. Seems silly to limit the tools used to analyze the question to what Einstein did. Physics is bigger than that.

[Mordred]

And where in his analysis of circular motion did Einstein use rapidity? I didn't see you make an analysis that allows an exact prediction of the time difference according to the ground frame, starting from the initial time difference D.

[edit:] Maybe you accept Einstein's result to which I linked in post #3, and which is easy to apply to the two clocks. Starting from that accepted result, calculating backwards (straight line constant speed) you should find the same as me.

 

And as you did not make clear with which step you have problems, let's take it step by step: please clarify the perceived error in my step 1.

 

 

I gave you an exact prediction example that you failed to include in this post.

 

[latex]F_a=F_b=\frac{mMG}{r^2}[/latex]

let x be the distance between a and b, let [latex]\alpha[/latex] represent the angle between one test body and the center line (center of gravity vertical axis) so from frame a, b experiences a force directed toward a.

[latex]F=2F_asin\alpha=2F_a*\frac{x}{2r}=\frac{mMG}{r^3}x[/latex]

a then observes b to be accelerating towards him by

[latex]F=-md^2x/dt^2[/latex]

[latex]\frac{d^2x}{dt^2}=\frac{MG}{r^3}x[/latex]

the 1/r^3 is characteristic of tidal forces.

That clears up the principle of equivalence a bit.... the key importance is to understand when this relation holds true.

the next part will take me a bit to type in...

the twin paradox which isn't a paradox at all, essentially can be broken down to the following statement.

"we should never have considered the age of a and b to be the same, as the frame of b undergoes accelerations that a does not undergo."

so lets look at the acceleration

first define the four velocity. [latex]u^\mu[/latex]

[latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex]

this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex]

the four velocity has constant length.

[latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex]

the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex]

[latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex]

so the acceleration and velocity four vectors are

[latex]c \frac{dt}{d\tau}=u^0[/latex]

[latex]\frac{dx^1}{d\tau}=u^1[/latex]

[latex]\frac{du^0}{d\tau}=a^0[/latex]

[latex]\frac{du^1}{d\tau}=a^1[/latex]

both vectors has vanishing 2 and 3 components.

using the equations above

[latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex]

in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here?

the last equation defines constant acceleration g. with solutions

[latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex]

from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex]

hence

[latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex]

similarly

[latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex]

so the solution to the last equation is

[latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex]

hence

[latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex]

with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex]

so

[latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

hence

[latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex]

and finally

[latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex],

[latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex]

the space and time coordinates then fall onto the Hyperbola during rotation

[latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex]

 

what did you think the last equation is? though commonly g is replaced with [latex]\propto^2[/latex] You can run the series of infinitesimals from these equations to plot your curved path. I'm certainly not going to waste my time doing it for you....

 

[latex]x^2-c^2\tau^2=\frac{c^4}{\propto^2}[/latex] as m_i=m_g principle of equivalence a change in velocity induces a rapidity rotation. Doesn't matter if its directional or the scalar component of the vector.

 

at another time I even posted you the action principle which correlates to this in terms of action. Specifically supplying and explaining the Principle of least action.

Edited October 10, 2016 by Mordred

[Tim88]

In the OP I ask to comment on my analysis, which I split out in 4 steps for targeted criticism. In the thereupon following posts I explain how my analysis follows the linked analyses of Einstein and Bell. No rapidity or further transformations are required in any of these cases.

 

Come on guys, this is very easy - especially with the already worked out examples:

 

- the effect of circular motion on clock rate was already explained by Einstein; that clarifies step 2.

- the method of comparing the identical motion of two objects that follow each other was already explained by Bell; that clarifies step 3.

- even the end result for one "moving" clock is given by Einstein, so that instead of following me, you can simply calculate backwards from the easily deduced results for two clocks that follow each other.

 

[edit:] Mordred I see that you added your unnecessarily complex calculation - it's like using a supercomputer simulation for adding groceries. As warned at the outset, it may be much more difficult to find the error in such complexity, regretfully. And the required answer is even missing - I even don't see any mention of the two car clocks!

 

Where do you show that the car clocks C1 and C2 need resynchronization due to the turnaround in order to make the one way speed of light again c for the car, and by how much?

 

Following my analysis, next the correct clock retardation values are found for each clock.

 

As a reminder, we are discussing here the measurement of the one way speed of light inside the car according to the car's clocks C1 and C2 and the ruler on the floor of the car. According to my analysis, for this particular case no synchronization correction is needed after turnaround in order to make the 1-way speed of light again c according to the car.

I suddenly notice what seems a disagreement with the referenced paper, according to which a clock at constant speed in a circle keeps the same rate as when going straight:

 

[..] different points on the car itself are undergoing spatial rotation. This changes the time interval due to rotation. Try and distinquish between "time" and "time interval".[..]

 

Note that my analysis is immune to that, as C1 and C2 undergo the identical effect from moving along the same circle segment.

Could it be that there is an ambiguity in the scenario, so that Mordred is trying to work out an unnecessary complex scenario while I chose on purpose one of the simplest possible scenarios? Anyone, is there a possible misunderstanding here??

Edited October 10, 2016 by Tim88

[Mordred]

Groan you do that under the 4 velocity section..... look at the statement about constant length. When you have rapidity=(velocity change) the length of the measuring rod has changed. See the section under acceleration and four vectors. OK visual aid time. Draw a line between two events. that line ds^2 is your shortest path at a specified (constant) velocity. An acceleration is a change in velocity. (plain and simple). Your path is no longer the same.

 

ds^2 is your worldline of simultaneous events for a specified velocity. When you change velocity your worldline is a different path altogether. Your Lorentz tranforms has a specific ds^2 line element. With your car scenario you have a Lorentz boost. However that Lorentz boost does not include rapidity... To have rapidity you need to use the Lorentz+rapidity equations. See under the following link.

https://en.wikipedia.org/wiki/Lorentz_transformation#boost

 

Here is the physical implications of rapidity. The last equation I posted is a simple correlation following the steps above. The original post those equations was on that you never linked here specified that those equations are in Lewis Ryder's Relativity book.

Edited October 10, 2016 by Mordred

[Tim88]

Mordred, that's great, you finally came to the point: your issue is with step 2.

The distance between the clocks doesn't stay exactly constant under rotation due to the rotation speed. Right?

Why didn't you say so immediately?

 

Indeed, that adds some more complexity to the analysis, which I had not included.

[edit: moved some later added text to the next post]

Edited October 10, 2016 by Tim88

[Mordred]

Mordred, that's great, you finally came to the point: your issue is with step 2.

The length doesn't stay exactly constant under rotation due to the rotation speed. Right? Why didn't you say so immediately?

under velocity change both change in speed or direction. I certainly tried to lol. Rotation speed isn't quite correct though, rapidity boosts is modeled via group rotation in much the same manner Lorentz boosts are modeled as rotations. You really have to spend considerable time studying the two to get a good picture. The best way is to reduce to the composition of velocities. Going through the step by steps to group aspects is too lengthy to cover here.

 

+1 on last post

Edited October 10, 2016 by Mordred

[Tim88]

I studied many years ago a related issue, the so-called double Fizeau toothed wheel. The axle twists due to the rotation speed added to the linear speed, in principle exactly by the right amount for the PoR to be maintained.

 

Still, the intention of the full example should not be overlooked, which is to keep it simple, neglecting minor effects, just as in the usual twin paradox a lot of approximations are made. How minor this one is still has to be analyzed, but it can perhaps in principle be compensated by some other imperfections such as the already mentioned asymmetry of the car due to the steering wheels. To be pondered over.

Edited October 10, 2016 by Tim88

[Tim88]

The devil is always in the details - and so we now have (top view, bend to the left), still ignoring dynamic effects:

 

1. The asymmetry of the car, which we chose to neglect and which can be compensated for by adding rear wheel steering:

 

----------

¦ ↑ ¦

vr ← \ C1 \

¦ ¦

¦ ↑ ¦

¦ C2 ¦

----------

 

I sketched the car almost normal length, else it becomes difficult to picture.

 

Without compensation of front wheel steering, C1 has a greater speed than C2 during the turnaround.

 

2. The additional inhomogeneous contraction of the car due to the composition of velocities (here sketched with full compensation of effect 1):

 

← ← ----------

¦ ↑ ¦

vr ← \ C1 \

¦ ↑ ¦

/ C2 / vr →

--------- → →

 

If I see it correctly, there will also be a skew which I did not sketch, but which doesn't matter for the trajectories for this symmetrical case.

 

This effect reduces the distance between C1 and C2.

 

As a result, C1 temporarily has a lower speed than C2 when going into the turn, and going out C1 temporarily has a greater speed than C2.

There is leeway for the driver to adapt the central speed of the car so as to keep for example the speed of C1 constant, or the speed of C2 constant, or to keep the speed of some other point of the car constant.

 

The original question has thus been refined to the question if in principle a combination of 1. and 2. is possible such that the result for time dilation is the same as when we just neglect them altogether - that would be neat (and it was the intention of this scenario that we may neglect such effects).

Edited October 11, 2016 by Tim88

[Mordred]

Average it as a bundle of worldlines for the car. World tube If I'm not mistaken is applicable. Note even a pointlike particle will undergo rapidity if it accelerates.

Of course we could ignore the car dimensions altogether and just use the centre which reduces the problem to one worldline. However that worldline still changes under rapidity.

The worldline calcs that includes rapidity is the Lorentz+rapidity equations. Those calcs assume a pointlike event. Where as the car is a collection of events.

I think perhaps you may be under a misconception of what rapidity entails in terms of events. An event is simply any pointlike coordinate of an pointlike object. Any additional coordinates used to define that object such as your car is a collection of coordinates. (a collection of events. Each with their own worldline.)

Within the volume of said car it is a field of events. Which can be modelled as an internal vector field.

 

This seems to be common problem. People first approaching relativity are too attached to materialistic views. I'd like to believe your beyond that, but I'm not positive you are.

 

For example "Is a particle solid ? or does it just display bullet like characteristics? as well as its wavelength.

Edited October 11, 2016 by Mordred

[Tim88]

I'm now working towards a reasonable "fix" so as to keep the example simple without brushing things under the carpet.

I'll post more later; first a better summary of points to consider, and next a pragmatic solution (I think, for it's "in progress").

[Tim88]

I noticed that I overlooked to include another classical effect.
When taking a turn (now assuming 4 wheel steering), and if we want to keep the speed of C1 and C2 constant (according to the ground frame), it means that their tangential speeds v_T = ds/dt remains constant along the circle segment. Consequently C1 should remain ahead of C2 by a distance of 1m along the circle segment. In order to achieve that, the car has to be made to contract - or at least, the distance between the clocks has to be reduced for the duration of the turn.

Let's see what we are dealing with here, this is simple geometry.

1.Classical effect.

Say we have the wheels of the car at 1m distance, beside the clocks, and we choose a crazy sharp bend of 100m radius.

I then find an angle of 2*0.28648° for a circle segment of about 1.000001m. Thus the car driver should ideally reduce the distance by about 1 micron (note that non-classically, that's 0.5 um in the ground frame for the chosen speed).

2. Now for the length contraction effect.

In view of the above, I now look differently at the issue of point 2 of post #15.
Here's my retake of it.

 

In principle an adapted 4WD and 4WS car with built-in flex can be conceived that keeps the clocks at constant speed.

We make the 2 clocks continue at a transverse speed of gamma=2 while on the circular trajectory. Correcting at the right time the distance by 1 um this can be reached, but now the middle of the car has a slightly lower speed as it's on the inside of the curvature.
Then the car floor with the ruler between the clocks will have the tendency to decontract. Here's an estimation of order of magnitude:
The middle of the car describes a circle trajectory with r=99.99875m.
Thus the transverse speed is 0.9999875*0.866025c = 0.866015c
If the whole ruler had that speed, its length would now increase to 0.500018m.
However it's somewhat in between, approximately 10 um, or 10x the classical effect.

In other words if I'm not mistaken, then small imperfections are strongly enhanced for this speed. That considerably complicates analysis.

 

But that's not all. Although there will not be a twist as in the double Focault disc as here all velocity vectors lie in the same plane, the right hand side of the car will contract more and the left hand side will decontract more. That causes a bending due to relativistic effects contrary to that of the bend. And that in turn leads to an increase of the decontraction issue.

 

And that's still not all, as inertia has so far been ignored. From the rotating disc I think to remember that inertial effects are larger than contraction, IOW, the decontraction issue may well turn into an additional contraction issue in the real world.

Although a specially built car was my first idea of how to fix this "problem within a problem", and it's not bad to maintain that, it's a unrealistic due to the extremely unrealistic speed. But that speed was only chosen for a clear pictorial illustration using gamma = 2. With that speed the car would be in 1s already far away from the Earth! Then we should include the loss of contact with the ground, the varying gravitational potential etc, all for just nonsense.

It's much more reasonable to instead discuss a technically possible scenario with a car going at for example 540 or 1080 km/h, and that should allow for greater analytical simplicity. Only disadvantage: it requires a calculator with many digits.

Edited October 14, 2016 by Tim88

[Tim88]

OK here are my estimations for the realistic section of the full car example.

If someone notices a significant error, please correct.

1080 km/h = 300 m/s => v/c = 1E-6
In good approximation gamma = 1+0.5*v²/c² => gamma = 1 + 5E-13

The initial clock (re-)synchronization of C1 and C2 in the car according to the ground proceeds as follows (length contraction can be neglected for this calculation, and we'll assume that the acceleration from stand still was gradual and smooth):

A light or radio pulse from C1 to C2 is clocked as 1/(3E8+300) m/s
= 3.3333300 ns, and in the inverse direction it's 3.3333367 ns.
This is very hard to detect, but apparently already possible - Chung 2003, Real-Time Detection of Femtosecond Optical Pulse Sequences (Freely downloadable, but the link may be illegal)

The driver will advance C2 by about 0.0000033ns = 3.3fs, so that the one-way speed of light becomes isotropically c in the car's newly established inertial reference system.

According to the ground's reckoning, the driver will now drop the ball in the front 3.3fs later than the ball in the rear; as a result the ball in the front will end up 300m/s * 3.3E-15s = 1E-12 m = 1pm too far, spanning a ground distance of 1m + 0.5pm (here's where the car's length contraction must be accounted for). If this were measurable, it would look for the car driver as if the ground is contracted by gamma = 1 + 5E-13.

Then the bend.
As expected, now the length contraction is negligible compared to the classical effect of 1 um "track expansion" on 1m clock distance as described in the foregoing post; it's even small compared to an atom!

And for 0.5 m to the outside of a r=100m bend, gamma = 1 + 5.5E-13. No surprises.

With a smooth 4 wheel in-steering from 0 to 100 m radius as discussed above, the middle point of the car has a speed of v_m= 0.9999875v so that delta-gamma = 0.5[(v/c)²-(v_m/c)²] = 1.25E-17.

10^-17 is negligible compared to 10^-11. On top of that, it's for a short duration and next the almost perfectly inverse effect is induced by smooth out-steering. Moreover, at such a high speed the turn radius should be well over 1km (ca. 3s of turning) instead of 100m.

Anyway, for simplicity we will assume that the car has a 4WD mechanism that literally forces C1 and C2 to keep exactly the same speed. For all practical purpose, C1 and C2 will then keep the same speed of 1080 km/h - even in the curve.
C1 and C2 are thus synchronized to "the car's inertial frame" on the outgoing leg as well as on the ingoing leg.

[edit: in practice this may be hard to achieve, due to vibrations, deformations, and imperfect clocks.]

To finish the example, we'll let the car do a return trip of 2*30km, or 60km in total.
We may neglect the rotation of the ground in this case.

The "twin" prediction of clock retardation on a clock that is located at the start/finish line, is then 5E-13 * 200s = 0.1 ns for C1 as well as for C2. That is also just about measurable.

 

As a reminder, the clocks follow an identical v(t) trajectory, so that they must accumulate the same retardations at the events of meeting up with the "stay-at-home" clock. The constant 3.3fs offset of C2 on C1 is independent of that fact, and there is no need to synchronize clocks to a ground clock: stopwatch times suffice for the comparisons.

 

In the course of 10 laps, C1 and C2 will accumulate 1.0 ns retardation.

Edited October 15, 2016 by Tim88

[Tim88]

Addendum: it will be useful to clarify that the driver will advance C2 by slightly more than 3.3fs, for initially we assume, as is usually done, that the distance between the clocks (as estimated from the ground) is allowed to relax after acceleration into a length contracted state so that C2 must be allowed to accelerate slightly more than C1. As here we are dealing with a car instead of a rocket, already at this point technical means should be provided for to make this approximately true. Alternatively the car driver could recalibrate the distance between the balls (but it's probably not yet technically measurable).

 

And to make it again more realistic, we better choose the wheel basis 3m and we can choose the radius of most of the turn for example 3km. The proper distance between the balls we now choose to be 2m (it's not necessary for the holes to be exactly between the wheels).

 

Then the advance of C2 on C1, as estimated from the ground, will be 6.7fs.

 

According to the ground's reckoning, the driver will now drop the ball in the front 6.7fs later than the ball in the rear; as a result the ball in the front will end up 300m/s * 6.7E-15s = 2E-12 m = 2pm too far, "spanning" a ground distance of 2m + 1pm (here's where the car's length contraction of 1pm matters).

[Mordred]

When I get time enough to be properly focussed. I'll look through it. Busy week

Edited October 16, 2016 by Mordred